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Can the mean of a frequency distribution be regarded as a weighted mean with relative frequencies as weights ?

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Surely. Let's imagine $j$ distinct values $x_j$ with frequency weights $w_j$. Then a weighted mean is $\sum_j w_j x_j / \sum_j w_j$.

Alternatively we can just write down all the values, say $n$ in number, whether distinct or not, so each has frequency $1$. The mean is then $(x_1 + \cdots + x_n)/n$, which we could redundantly write $(1\ x_1 + \cdots + 1\ x_n) / n$. This is exactly equivalent to weights $w_i$ all $1$, yielding expressions that all mean the same for frequencies of $1$: $$\sum_{i=1}^n w_i x_i\,/\,\sum_{i=1}^n w_i\ =\ \sum_{i=1}^n w_i x_i\, /\, n\ =\ (\sum_{i=1}^n 1\ x_i)\, /\, n = \sum_{i=1}^n x_i\, /\, n.$$

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